3.43 \(\int \frac{\sec ^4(c+d x)}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=85 \[ -\frac{2 \tan (c+d x)}{a d}+\frac{3 \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{\tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac{3 \tan (c+d x) \sec (c+d x)}{2 a d} \]

[Out]

(3*ArcTanh[Sin[c + d*x]])/(2*a*d) - (2*Tan[c + d*x])/(a*d) + (3*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (Sec[c +
d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x]))

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Rubi [A]  time = 0.0927384, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3818, 3787, 3767, 8, 3768, 3770} \[ -\frac{2 \tan (c+d x)}{a d}+\frac{3 \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{\tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac{3 \tan (c+d x) \sec (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + a*Sec[c + d*x]),x]

[Out]

(3*ArcTanh[Sin[c + d*x]])/(2*a*d) - (2*Tan[c + d*x])/(a*d) + (3*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (Sec[c +
d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x]))

Rule 3818

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(d^2*Cot[e
+ f*x]*(d*Csc[e + f*x])^(n - 2))/(f*(a + b*Csc[e + f*x])), x] - Dist[d^2/(a*b), Int[(d*Csc[e + f*x])^(n - 2)*(
b*(n - 2) - a*(n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x)}{a+a \sec (c+d x)} \, dx &=-\frac{\sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac{\int \sec ^2(c+d x) (2 a-3 a \sec (c+d x)) \, dx}{a^2}\\ &=-\frac{\sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac{2 \int \sec ^2(c+d x) \, dx}{a}+\frac{3 \int \sec ^3(c+d x) \, dx}{a}\\ &=\frac{3 \sec (c+d x) \tan (c+d x)}{2 a d}-\frac{\sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac{3 \int \sec (c+d x) \, dx}{2 a}+\frac{2 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{a d}\\ &=\frac{3 \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{2 \tan (c+d x)}{a d}+\frac{3 \sec (c+d x) \tan (c+d x)}{2 a d}-\frac{\sec ^2(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}\\ \end{align*}

Mathematica [B]  time = 1.33355, size = 250, normalized size = 2.94 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos \left (\frac{1}{2} (c+d x)\right ) \left (-\frac{4 \sin (d x)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{1}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{1}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-6 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-4 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )\right )}{2 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + a*Sec[c + d*x]),x]

[Out]

(Cos[(c + d*x)/2]*Sec[c + d*x]*(-4*Sec[c/2]*Sin[(d*x)/2] + Cos[(c + d*x)/2]*(-6*Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] + 6*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^(-2) - (Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])^(-2) - (4*Sin[d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)
/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))))/(2*a*d*(1 + Sec[c + d*x]))

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Maple [A]  time = 0.039, size = 143, normalized size = 1.7 \begin{align*} -{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{3}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{3}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{3}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{3}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+a*sec(d*x+c)),x)

[Out]

-1/a/d*tan(1/2*d*x+1/2*c)-1/2/a/d/(tan(1/2*d*x+1/2*c)+1)^2+3/2/a/d/(tan(1/2*d*x+1/2*c)+1)+3/2/a/d*ln(tan(1/2*d
*x+1/2*c)+1)+1/2/a/d/(tan(1/2*d*x+1/2*c)-1)^2+3/2/a/d/(tan(1/2*d*x+1/2*c)-1)-3/2/a/d*ln(tan(1/2*d*x+1/2*c)-1)

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Maxima [B]  time = 1.14508, size = 219, normalized size = 2.58 \begin{align*} -\frac{\frac{2 \,{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac{2 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos
(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 3*l
og(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 2*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.95933, size = 301, normalized size = 3.54 \begin{align*} \frac{3 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{4 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(3*(cos(d*x + c)^3 + cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 3*(cos(d*x + c)^3 + cos(d*x + c)^2)*log(-sin(
d*x + c) + 1) - 2*(4*cos(d*x + c)^2 + cos(d*x + c) - 1)*sin(d*x + c))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2
)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{4}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+a*sec(d*x+c)),x)

[Out]

Integral(sec(c + d*x)**4/(sec(c + d*x) + 1), x)/a

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Giac [A]  time = 1.42713, size = 136, normalized size = 1.6 \begin{align*} \frac{\frac{3 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{3 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} + \frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

1/2*(3*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 3*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a - 2*tan(1/2*d*x + 1/2*c)/
a + 2*(3*tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a))/d